Problem 71 - University Physics with Modern Physics

Question

CALC An object is attracted toward the origin with a force given by \(F_{x}=-k / x^{2}\) . (Gravitational and electrical forces have this distance dependence.) (a) Calculate the work done by the force \(F_{x}\) when the object moves in the \(x\) -direction from \(x_{1}\) to \(x_{2}\) . If \(x_{2}>x_{1},\) is the work done by \(F_{x}\) positive or negative? (b) The only other force acting on the object is a force that you exert with your hand to move the object slowly from \(x_{1}\) to \(x_{2} .\) How much work do you do? If \(x_{2}>x_{1},\) is the work you do positive or negative? (c) Explain the similarities and differences between your answers to parts (a) and (b).

Step-by-Step Solution

Answer

(a) The work done by \( F_x \) is \( \frac{k}{x_1} - \frac{k}{x_2} \) and is positive. (b) The work you do is \( \frac{k}{x_2} - \frac{k}{x_1} \) and is negative. (c) Both works are equal in magnitude but opposite in sign.

Step 1: Understanding the Work Done by Force Fx

The work done by a force when an object moves is the integral of the force over the distance. Here, the force is given by \( F_{x} = -\frac{k}{x^2} \). The object moves from \( x_1 \) to \( x_2 \), thus the work done by the force \( F_{x} \) is:\[ W = \int_{x_1}^{x_2} F_x \, dx = \int_{x_1}^{x_2} -\frac{k}{x^2} \, dx \]

Step 2: Integrating the Force Expression

The integral of \( -\frac{k}{x^2} \) with respect to \( x \) is calculated as follows:\[ \int -\frac{k}{x^2} \, dx = k \int x^{-2} \, dx = k \left[-x^{-1}\right] = -\frac{k}{x} \] Now apply the limits from \( x_1 \) to \( x_2 \):\[ W = \left.-\frac{k}{x}\right|_{x_1}^{x_2} = -\frac{k}{x_2} - \left(-\frac{k}{x_1}\right) = \frac{k}{x_1} - \frac{k}{x_2} \]

Step 3: Determining the Sign of Work Done by Fx

If \( x_2 > x_1 \), then \( \frac{k}{x_1} > \frac{k}{x_2} \) because \( \frac{k}{x} \) decreases as \( x \) increases. Thus, \( W = \frac{k}{x_1} - \frac{k}{x_2} \) is positive. Therefore, the work done by \( F_{x} \) is positive when \( x_2 > x_1 \).

Step 4: Work Done by Your Hand

The work done by the force you exert with your hand can be considered the negative of the work done by \( F_x \) because you are moving the object against the attractive force. Using the formula from Step 2:\[ W_{hand} = -\left(\frac{k}{x_1} - \frac{k}{x_2}\right) = \frac{k}{x_2} - \frac{k}{x_1} \]

Step 5: Determining the Sign of Work Done by Your Hand

Since \( x_2 > x_1 \), \( \frac{k}{x_2} < \frac{k}{x_1} \), making \( W_{hand} = \frac{k}{x_2} - \frac{k}{x_1} \) negative. Therefore, when \( x_2 > x_1 \), the work you do is negative.

Step 6: Comparing Work Done by Fx and Your Hand

The work done by the force \( F_x \) is positive, indicating energy is extracted from the system, while the work done by your hand is negative, implying energy is input into the system. This reflects the balance of forces as you exert work against the natural force acting on the object.

Key Concepts

Force Integration

When calculating the work done by a force, we use an integration process. This involves the integral of the force over the specified distance. In this exercise, the force is given by the expression \( F_{x} = -\frac{k}{x^2} \).As the object moves from \( x_1 \) to \( x_2 \), the work done by this force is the integral:\[ W = \int_{x_1}^{x_2} F_x \, dx = \int_{x_1}^{x_2} -\frac{k}{x^2} \, dx \]**Integration Steps:**- Start by rewriting the force function for easier integration: \(-\frac{k}{x^2} = -k\cdot x^{-2} \).- Integrate with respect to \( x \): \[ \int -k\cdot x^{-2} \, dx = -k \int x^{-2} \, dx = k \left[-x^{-1}\right] = -\frac{k}{x} \].- Apply the limits from \( x_1 \) to \( x_2 \): \[ W = \left.-\frac{k}{x}\right|_{x_1}^{x_2} = \frac{k}{x_1} - \frac{k}{x_2} \].This process of integrating allows us to determine the total work done by the force over a continuous range, refining our understanding of how forces operate over distances.

Attractive Forces

Attractive forces such as gravitational and electrical forces pull objects closer together.
The force involved in this exercise, \( F_{x} = -\frac{k}{x^2} \), is typical of these forces.
Negative signs here indicate that the force is directed towards the source, which is often at an origin or a central point.**Characteristics of Attractive Forces:**- **Dependence on Distance:** The force's magnitude is inversely proportional to the square of the distance (\( x^2 \)).
This means it gets stronger pretty fast as you get closer to the origin.
- **Directional Nature:** These forces always aim to pull or attract the object towards a smaller distance, often denoted by "negative" work, particularly if external work is done to move against this force.
Understanding these properties helps explain how natural forces work, such as gravity or electrostatic forces among charged particles. This concept is essential in physics, as it explains how planets orbit stars and how electrons behave around nuclei.

Negative Work

Work done is classified into positive or negative, depending on the direction of force and motion. In cases of attractive forces, like seen in this exercise, the concept of negative work becomes important.**What is Negative Work?**- **Opposite Direction of Motion:** When a force acts against the direction of movement, negative work is done.
For example, if you're pushing an object up a slope while gravity pulls it down, you're doing negative work relative to gravity's direction.
- **Energy Considerations:** Negative work can mean that energy is being taken from the system, requiring external energy input to maintain motion.
In this problem, moving the object from \( x_1 \) to \( x_2 \) requires work by your hand against the natural force, resulting in negative work from your perspective. \[ W_{hand} = \frac{k}{x_2} - \frac{k}{x_1} \]The concept of negative work is crucial in various real-world applications, such as brakes in vehicles or energy absorption mechanisms. Understanding when and how negative work is done allows us to manage systems that require energy balance and compensation effectively.